How do you solve $x = \sqrt{- 7 x + 8}$ $x=sqrt( -7x+8)$ and find any extraneous solutions?

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Answer 1 · 2016-07-16T04:05:44

${(x)}^{2} = {(\sqrt{- 7 x + 8})}^{2}$ $(x)^2 = (sqrt(-7x + 8))^2$

$x^{2} = - 7 x + 8$ $x^2 = -7x + 8$

$x^{2} + 7 x - 8 = 0$ $x^2 + 7x - 8 = 0$

$(x + 8) (x - 1) = 0$ $(x + 8)(x - 1) = 0$

$x = - 8 and 1$ $x = -8 and 1$

Checking in the original équation, we find only $x = 1$ $x = 1$ works. Hence, ${1}$ ${1}$ is the solution set.

Hopefully this helps!

How do you solve x=√−7x+8x=sqrt( -7x+8) and find any extraneous solutions?